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What does birational equivalence signify geometrically?

I have been reading AG and my question is why birational equivalence came into the picture? My basic intuition is we may study a space, $X$ not so much different from the space, $Y$ and we will still be able to absorb great amount of information of the space, $Y$. Because except at certain points they are almost similar(?). For example the affine line and the hyperbola are birational. But! What information in general two birational objects posses?

I have just started reading the subject. Can anyone please explain it to me or provide me some resource?

nkh99
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1 Answers1

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One important fact towards understanding birational maps is the following, which appears for example in Hartshorne as Lemma V.5.1.

Let $f: X \dashrightarrow Y$ be a rational map of projective varieties, with $X$ normal. Then, the exceptional set $\operatorname{Ex}(f)$ has codimension $\geq 2$ in $X$.

Note for example that this implies that a rational map from a smooth projective curve is always regular. This can be seen in more concrete ways, though. Below I record an example of how this fact can be used to deduce birational invariants of smooth algebraic varieties.

Let $X$ be a smooth projective variety over $\mathbb{C}$. Denote $\Omega_{X}$ the sheaf of algebraic $1$-forms on $X$ and $\omega_X = \det \Omega_X$ its top exterior power. This line bundle is often called the canonical bundle of $X$. Being a line bundle, there is an induced map $\varphi: X \dashrightarrow \mathbb{P}(\Gamma(X, \omega_X)^{\vee})$.

(Explicitly $\varphi$ sends $x \in X$ to the projective equivalence class of the evaluation $\operatorname{ev}_x: \Gamma(X, \omega_X) \to \mathbb{C}$. Note that this is only well defined when $\operatorname{ev}_x$ is not the zero functional: not all global sections of $\omega_X$ vanish at $x$.)

We call $\varphi$ the first canonical map and the (closure of) the image the first canonical image $X^{[1]} \subset \mathbb{P}(\Gamma(X, \omega_X)^\vee)$. As the name and notation suggest, one can repeat this with $\omega_X^{\otimes n}$ to obtain the $n$'th canonical map.

I claim that all of these are birational invariants. Let $f: X \dashrightarrow Y$ denote a birational map of smooth varieties.

Indeed, if $U \subset X$ is the complement of $\operatorname{Ex}(f)$ then there is a birational map $f|_U: U \to Y$ which induces a map $f^*: \Gamma(Y, \omega_Y^{\otimes n}) \to \Gamma(U, \omega_U^{\otimes n})$. But then $\omega_U^{\otimes n} \cong (\omega_X^{\otimes n})|_U$ and since $U$ is the complement of a closed set of codimension $\geq 2$, $\Gamma(U, \omega_U^{\otimes n}) = \Gamma(X, \omega_X^{\otimes n})$.

Hence there is an induced map $f^*: \Gamma(Y, \omega_Y^{\otimes n}) \to \Gamma(X, \omega_X^{\otimes n})$. Symmetrically one can define $(f^{-1})^*: \Gamma(X, \omega_{X}^{\otimes n}) \to \Gamma(X, \omega_Y^{\otimes n})$ and one can check that this is in fact an inverse to $f^*$.

This then induces a linear isomorphism $\mathbb{P}(\Gamma(X, \omega_X^{\otimes n})^\vee) \cong \mathbb{P}(\Gamma(Y, \omega_Y^{\otimes n})^\vee)$ which identifies $X^{[n]}$ with $Y^{[n]}$, up to a linear automorphism of $\mathbb{P}(\Gamma(Y, \omega_Y^{\otimes n})^\vee)$.

Hence to every birational equivalence class of varieties one can associate a sequence of projective varieties $X^{[n]}$ and in fact, for $n$ large and divisible these varieties stabilize to a variety $X^{\text{stab}}$ which we call the stable canonical image. Since the canonical line bundle is the only natural nontrivial line bundle to consider on any smooth variety, this is a rather strong constraint on any given birational equivalence class.

A good elementary overview of this kind of thing can be found in Koll'ar's paper The Structure of Algebraic Threefolds.

Daniel
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